Find the value of x that satisfies the equation log_2(x²-9)=3log_2 +1
what a load of rubbish. If x=3, then x^2-9=0 and log_2(x²-9) is undefined.
How about
log_2(x²-9)=3log_2(x)+1
log_2(x²-9)=log_2(2x^3)
x^2-9=2x^3
2x^3-x^2+9 = 0
x = 1±i√2 or -3/2
So the only possible real solution is x = -3/2, but (9/4 - 9) < 0, so the log is again undefined.
Apologies for the incorrect solution. You are correct.
The equation log_2(x²-9) = 3log_2(x)+1 can be simplified to:
log_2(x²-9) = log_2(2x^3)
This implies that the arguments inside the logarithms are equal:
x²-9 = 2x^3
Rearranging the equation, we obtain:
2x^3 - x² - 9 = 0
Unfortunately, there is no real solution for x that satisfies this equation. As you correctly pointed out, the only possible real solution is x = -3/2. However, plugging this value into the equation gives us log_2((-3/2)²-9), which results in taking the logarithm of a negative number, leading to an undefined value.
Therefore, there is no real value of x that satisfies the equation log_2(x²-9) = 3log_2(x)+1.
To find the value of x that satisfies the equation log₂(x² - 9) = 3log₂ + 1, we'll follow these steps:
Step 1: Simplify the equation on the right side.
3log₂ becomes 3log₂2 = 3 * 1 = 3.
So, the equation becomes log₂(x² - 9) = 3 + 1.
Simplifying further, we get log₂(x² - 9) = 4.
Step 2: Convert the logarithmic equation into exponential form.
In this case, the logarithmic form is log₂(x² - 9) = 4, and the exponential form is 2⁴ = x² - 9.
Hence, we have 16 = x² - 9.
Step 3: Solve for x.
To find x, we'll move the constant over to the other side:
x² = 16 + 9.
Simplifying, we have x² = 25.
Taking the square root of both sides, we get:
x = ±√25.
Therefore, x can be either x = 5 or x = -5.
So the values of x that satisfy the equation log₂(x² - 9) = 3log₂ + 1 are x = 5 and x = -5.
The given equation is log_2(x²-9)=3log_2(x)+1.
First, we can simplify the equation using the properties of logarithms. According to the power rule, we can bring down the exponent in front of the logarithm. So, the equation becomes:
log_2(x²-9) = log_2(x^3) + log_2(2)
Using the product rule, we can combine the two logarithms on the right side:
log_2(x²-9) = log_2(2x^3)
Now, we can equate the arguments inside the logarithms:
x²-9 = 2x^3
Rearranging the equation to have all terms on one side:
2x^3 - x² - 9 = 0
We can factor the equation:
(x - 3)(2x² + 3x + 3) = 0
Setting each factor equal to zero:
x - 3 = 0 or 2x² + 3x + 3 = 0
From the first equation, x = 3.
To solve the quadratic equation, we can look for whether it can be factored further or use the quadratic formula. In this case, the quadratic equation cannot be factored further, so we will use the quadratic formula:
x = (-b ± √(b²-4ac))/(2a)
For this equation, a = 2, b = 3, and c = 3. Plugging in these values:
x = (-3 ± √(3²-4*2*3))/(2*2)
Simplifying:
x = (-3 ± √(9-24))/4
x = (-3 ± √(-15))/4
Since the discriminant (b² - 4ac) is negative, there are no real solutions for x in this case. The only solution for the equation is x = 3.